Let `f:N->N` be defined by `f(x)=x^2+x+1,x in N`. Then is `f` is

Options

A. one-one and onto
B. many-one onto
C. one-one but not onto
D. none of these

Answer

C

Explanation

Let `x,y in N` such that `f(x)=f(y).` Then
`f(x)=f(y)`
or `x^(2)+x+1=y^(2)+y+1`
or `(x-y)(x+y+1)=0`
i.e., `x=y " or " x=(-y-1) notin N`
Therefore, `f` is one-one.
Also, `f(x)` does not take all positive integral values. Hence, `f` is into.